What will be the sollution if

Question

$$\begin{align*} 7&=7 \\ 7\cdot7&=49 \\ 7\cdot7\cdot7&=343 \\ 7\cdot7\cdot7\cdot7&=2041\end{align*}$$ What will be the last digit if $111$ number of $7$s are multiplied?

Answer 1

Just touching on the hint that I gave you in the comments:

Since $7$ and $10$ have greatest common divisor $1$; we have $7^{\phi(10)}\equiv 1\bmod 10$, that is $7^{4}\equiv1\bmod 10$. Then $$(7^{4})^{27}=7^{108}\equiv1\bmod10\Rightarrow7^{111}=7^{108}\cdot7^{3}=7^{3}\equiv3\bmod10.$$

Alternatively, brute force for a few terms will give that the powers of $7$ have last digit following a periodic pattern of $7,9,3,1$. Since $111$ is $3\bmod 4$; it follows that the last digit will be $3$.

Answer 2

Hint $$7^1={\bf 7} \qquad 7^5=1680{\bf 7} \\ 7^2=4{\bf 9} \qquad 7^6=11764{\bf 9}\\ 7^3=34{\bf 3} \qquad ...\\ 7^4=204{\bf 1} \qquad ...\\ $$